Interval for area bounded by $r = 1 + 3 \sin \theta$

Question

I'm trying to calculate the area of the region bounded by one loop of the graph for the equation

$$ r = 1 + 3 \sin \theta $$

I first plot the graph as a limaçon with a maximum outer loop at $(4, \frac{\pi}{2})$ and a minimum inner loop at $(-2, -\frac{3 \pi}{2})$. I then note the graph is symmetric with respect to the $\frac{\pi}{2}$ axis and the zero for the right half is at $\theta = \arcsin(-\frac{1}{3})$.

So, I chose the interval $[\arcsin(-\frac{1}{3}),\frac{\pi}{2}]$ to calculate the area which can then be multiplied by $2$ for the other half. The problem is that the answer in the book seems to use $\arcsin(\frac{1}{3})$ instead, note the change of sign.

Just to make sure I'm not misunderstanding where I went wrong, I get the answer

$$ \frac{11 \pi}{4} - \frac{11}{2} \arcsin(-\frac{1}{3}) + 3 \sqrt 2 $$

Whereas the book gets

$$ \frac{11 \pi}{4} - \frac{11}{2} \arcsin(\frac{1}{3}) - 3 \sqrt 2 $$

It's a subtle change of sign but I'd really like to understand where I went wrong.

Answer 1

Notice how $\arcsin(-\frac{1}{3}) = - \arcsin(\frac{1}{3})$, so your answer now looks like

$$ \frac{11 \pi}{4} + \frac{11}{2} \arcsin(\frac{1}{3}) + 3 \sqrt 2 \\ $$

That means your area is greater than the answer in your book by:

$$ 2 \left(\frac{11}{2} \arcsin(\frac{1}{3}) + 3 \sqrt 2\right) $$

This might indicate you are calculating the area of the outer loop whereas your book is calculating the inner loop. If you choose the interval $[\frac{3 \pi}{2}, 2 \pi - \arcsin(\frac{1}{3})]$ to calculate the half as you did before, you get:

$$ \begin{eqnarray} A &=& 2 \times \frac{1}{2} \int_{\frac{3 \pi}{2}}^{2 \pi - \arcsin \frac{1}{3}} (1 + 3 \sin \theta)^2 \, \textrm{d}\theta \\ &=& \left[\frac{11 \theta}{2} - 6 \cos \theta - \frac{9 \sin(2 \theta)}{4} \right]_{\frac{3 \pi}{2}}^{2 \pi - \arcsin \frac{1}{3}} \\ &=& \frac{11 \pi}{4} - \frac{11}{2} \arcsin(\frac{1}{3}) - 3 \sqrt 2 \\ \end{eqnarray} $$

This seems to agree with the answer in your book.

Answer 2

$$ \begin{align} \int_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\frac12r^2\,\mathrm{d}\theta &=\int_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\frac12(1+3\sin(\theta))^2\,\mathrm{d}\theta\\ &=\int_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\frac12\left(1+6\sin(\theta)+9\sin^2(\theta)\right)\mathrm{d}\theta\\ &=\int_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\frac12\left(1+6\sin(\theta)+9\left(\frac{1-\cos(2\theta)}2\right)\right)\mathrm{d}\theta\\ &=\left[\frac{11}4\theta-3\cos(\theta)-\frac98\sin(2\theta)\right]_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\\ &=\frac{11}4\left(\pi+2\arcsin\left(\frac13\right)\right)+3\sqrt2 \end{align} $$ This is the area of the outer loop

The sum of both loops is $$ \begin{align} &\int_0^{2\pi}\frac12\left(1+6\sin(\theta)+9\left(\frac{1-\cos(2\theta)}2\right)\right)\mathrm{d}\theta\\ &=\left[\frac{11}4\theta-3\cos(\theta)-\frac98\sin(2\theta)\right]_0^{2\pi}\\ &=\frac{11}2\pi \end{align} $$ so the area of the inner loop is $$ \frac{11}4\left(\pi-2\arcsin\left(\frac13\right)\right)-3\sqrt2 $$